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A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{~m} / \mathrm{s}^{2}\). (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{~s}\). Find the acceleration \(\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right)\) of the dragster.

Short Answer

Expert verified
(a) It takes 11.41 seconds for the Beetle. (b) The dragster's acceleration is 44.70 m/s².

Step by step solution

01

Convert Initial and Final Velocity Units

The Beetle's speed is given in miles per hour (mi/h). Convert this to meters per second (m/s) using the conversion factor: 1 mile = 1609.34 meters 1 hour = 3600 seconds. Thus, \(60.0 \text{ mi/h} = 60.0 \times \frac{1609.34}{3600} \approx 26.82 \text{ m/s} \).
02

Use the Formula for Time

Use the formula for acceleration, which is: \[ a = \frac{{v_f - v_i}}{t} \]where \( v_f = 26.82 \text{ m/s} \), \( v_i = 0 \text{ m/s} \), and \( a = 2.35 \text{ m/s}^2 \).Rearrange the formula to solve for time (\( t \): \[ t = \frac{{v_f - v_i}}{a} = \frac{{26.82 - 0}}{2.35} \approx 11.41 \text{ seconds} \].
03

Calculate the Dragster's Acceleration

For the dragster, we need to calculate the acceleration given time (\( t = 0.600 \text{ s} \)) and the same speed conversion for final velocity (\( v_f = 26.82 \text{ m/s} \)). The initial velocity is again 0.Use the formula: \[ a = \frac{{v_f - v_i}}{t} = \frac{{26.82 - 0}}{0.600} = 44.70 \text{ m/s}^2 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration
Acceleration is a fundamental concept in kinematics, which describes the change in velocity of an object over time. When you press the gas pedal in a car, the vehicle accelerates. This means there is an increase in speed. Acceleration can be both positive and negative, where positive acceleration means speeding up, and negative means slowing down.

The formula used to calculate acceleration is:
  • \( a = \frac{{v_f - v_i}}{t} \)
where:
  • \( a \) is the acceleration,
  • \( v_f \) is the final velocity,
  • \( v_i \) is the initial velocity,
  • \( t \) is the time period over which the change occurs.
In our example with the VW Beetle, we know the car's initial velocity \( v_i \) is 0 m/s and the final velocity \( v_f \) is calculated to be 26.82 m/s. The car accelerates at \( 2.35 \, \text{m/s}^2 \). You can see how the formula explains that the faster the velocity changes, the greater the acceleration.
Velocity Conversion
To analyze any problem in kinematics correctly, sometimes you need to convert units, especially when different units of measurements are used. Many times, speed is given in miles per hour (mi/h), but calculations often require meters per second (m/s) instead. The conversion helps keep the calculations consistent and easier to understand.

Here, the conversion process is critical. To convert 60.0 mi/h to m/s, use the conversion factors:
  • 1 mile = 1609.34 meters
  • 1 hour = 3600 seconds
The conversion would look like this:
  • \( 60.0 \, \text{mi/h} \times \frac{1609.34}{3600} = 26.82 \, \text{m/s} \)
Hence, converting velocity from mi/h to m/s ensures that all measures are in the same units, simplifying further calculations. Without converting, the application of the acceleration formula would be incorrect.
Calculating Time from Acceleration
The calculation of time when acceleration is known is handy. It allows us to determine how quickly an object gets to a certain velocity. In the scenario involving our VW Beetle, we use the equation for acceleration to solve for time. Rearranging the formula \( a = \frac{{v_f - v_i}}{t} \) to solve for \( t \) gives:
  • \( t = \frac{{v_f - v_i}}{a} \)

Please note:
  • \( v_f = 26.82 \, \text{m/s} \)
  • \( v_i = 0 \, \text{m/s} \)
  • \( a = 2.35 \, \text{m/s}^2 \)
By applying these values to the formula, the time \( t \) taken for the VW Beetle to reach its target speed is calculated as follows:
  • \( t = \frac{26.82}{2.35} \approx 11.41 \, \text{seconds} \)
Understanding how to manipulate this equation let us predict how long it takes to accelerate to a new speed. This application is valuable not just in physics education but also in practical, real-world scenarios autonomously, such as in sports or automotive testing.

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Most popular questions from this chapter

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